# An upper bound for the maximal subgroups at fixed index?

Let us call a subgroup an injective homomorphism between groups.

I warn the reader that a subgroup designates here an inclusion $(H \subset G)$, not $H$ alone.

A subgroup $H \subset G$ is maximal if for all intermediate subgroups $H \subset K \subset G$, then $K=H$ or $G$.

Let $\sim$ be the equivalence of subgroups (defined here).
Note that the maximality is invariant under $\sim$.

Let $n$ be a fixed integer. For each equivalence class of index $n$ maximal subgroups, we choose a representative $(H \subset G)$ with $G$ of minimal order. Let $R_{n}$ be the set of all these representatives.

Interrelated questions :

• Does $R_{n}$ is a finite set ?
• $\forall (H \subset G) \in R_{n}$, is $ord(G)$ bounded ?
• Is $G$ always a finite group ? A counter-example ?

Original motivation: What's the list of all the maximal subgroups at index $6$ ?

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Namely, let $H\subset G$ be of finite index $n$. Since the intersection $K$ of all conjugates of $H$ in $G$ has index dividing $n!$ in $G$ (it is the kernel of the obvious morphism $G\to S_n$) and is of course normal in $G$, you have $(H\subset G)\sim (H/K\subset G/K)$.
The case of maximal subgroups is then about primitive (in particular transitive) finite permutation subgroups of $S_n$.
BS: Your answer works, but only for finitely-generated groups. In the infinitely-generated case $Hom(G,S_n)$ could be infinite and an answer would require opening up the definition of "equivalence" in the OP. Another caveat is that, on the face of it, the OP uses very unusual definition of a "subgroup." With this definition, $Z^2$ contains infinitely many subgroups of index 1. Maybe this does not matter though (again, it depends on the definition of equivalence). – Misha Jul 9 '13 at 23:06